Description

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input:

board = [
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
words = ["oath","pea","eat","rain"]

Output:

["eat","oath"]

Note:

  1. All inputs are consist of lowercase letters a-z.
  2. The values of words are distinct.

分析

题目的意思是：给定一个二维字符矩阵，和一个单词列表，问有多少个单词能够由二维字符矩阵的一条路径构成，其中相同字符单元只能用一次。

  • 首先建立了一个trie的树，然后在这个字典树里面查找。

代码

class TrieNode{
public:
    int isWord;
    vector<TrieNode*> next;
    TrieNode(){
        isWord=-1;
        next=vector<TrieNode*>(26,NULL);
    }
};

class Solution {
private:
    TrieNode* root;
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        root=new TrieNode();
        buildTree(words,root);
        set<string> res;
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[i].size();j++){
                findWords(res,board,words,root,i,j);
            }
        }
        return vector<string>(res.begin(),res.end());
        
    }
    
    void findWords(set<string>& res, vector<vector<char>>& board, vector<string>& words, TrieNode* node, int r, int c){
        if(board[r][c] == '.' || !node->next[board[r][c]-'a']) return;
        TrieNode* after=node->next[board[r][c]-'a'];
        if(after->isWord>-1){
            res.insert(words[after->isWord]);
        }
        char letter=board[r][c];
        board[r][c]='.';
        if(r-1>=0){
            findWords(res,board,words,after,r-1,c);
        }
        if(r+1<board.size()){
            findWords(res,board,words,after,r+1,c);
        }
        if(c-1>=0){
            findWords(res,board,words,after,r,c-1);
        }
        if(c+1<board[0].size()){
            findWords(res,board,words,after,r,c+1);
        }
        board[r][c]=letter;
        
    }
    
    void buildTree(vector<string>& words,TrieNode* root){
        TrieNode* cur;
        for(int i=0;i<words.size();i++){
            cur=root;
            for(char c:words[i]){
                if(!cur->next[c-'a']){
                    cur->next[c-'a']=new TrieNode();
                }
                cur=cur->next[c-'a'];
            }
            cur->isWord=i;
        }
    }
};

参考文献

​​212. Word Search II​​